1. ## Prove identity

Prove that $\arctan x+\arctan y\equiv \arctan\frac{x+y}{1-xy}$.
use this relationship to show that, if $\arctan x+\arctan y+arctan z=\frac{\pi}{2}$ then $xy+yz+xz=1$
I have done the first part:
Let $\arctan x=\theta$ and $\arctan y=\phi$
then for LHS
$\equiv \theta +\phi$
$\equiv \arctan(\tan(\theta+\phi))$
$\equiv \arctan(\frac{tan \theta+\tan \phi}{1-\tan \theta \tan \phi})$
$\equiv \arctan(\frac{x+y}{1-xy})\equiv RHS$
now my problem is the second part, after i subsitute $\arctan(\frac{x+y}{1-xy})$ into $\arctan x+\arctan y+arctan z=\frac{\pi}{2}$ to get $\arctan(\frac{x+y}{1-xy})+\arctan z=\frac{\pi}{2}$, i don't know what to do.
Thanks

2. Let $\frac{x+y}{1-xy}=w$

Apply again the first identity for $\arctan w+\arctan z$ and replace back w.

3. thanks! i did that, and let $arctan w=\omega$ and $arctan z=\gamma$
so i had $\omega+\gamma=\frac{\pi}{2}$
$arctan(tan(\omega+\gamma)=\frac{\pi}{2}$
$arctan(\frac{w+z}{1-wz})=\frac{\pi}{2}$
$arctan(\frac{\frac{x+y}{1-xy}+z}{1-z(\frac{x+y}{1-xy})})=\frac{\pi}{2}$
$arctan(\frac{x+y+z-xyz}{1-xy-xz-yz})=\frac{\pi}{2}$
now i'm stuck, $tan\frac{\pi}{2}$ doesn't give a value

4. If $\arctan\frac{a}{b}=\frac{\pi}{2}$ then b must be 0.

In this case $1-xy-xz-yz=0\Rightarrow xy+xz+yz=1$

Edited.

5. i thinking something along that line but didn't have any confirmation
thanks!

6. Originally Posted by red_dog
If $\arctan\frac{a}{b}=0$ then b must be 0.

In this case $1-xy-xz-yz=0\Rightarrow xy+xz+yz=1$
Whoops! I think you mean 'If $\arctan\frac{a}{b}=\frac{\pi}{2}$ then b must be 0'.

Whoops! I think you mean 'If $\arctan\frac{a}{b}=\frac{\pi}{2}$ then b must be 0'.